Get Answers to all your Questions

header-bg qa

The probability that 4th power of a positive integer ends in the digit 6 is:

Option: 1

10\; ^{o}/_{o}
 


Option: 2

20\; ^{o}/_{o}


Option: 3

25\; ^{o}/_{o}

 


Option: 4

40\; ^{o}/_{o}


Answers (1)

best_answer

 

Probability of occurrence of an event -

Let S be the sample space then the probability of occurrence of an event E is denoted by P(E) and it is defined as 

P\left ( E \right )=\frac{n\left ( E \right )}{n\left ( S \right )}

P\left ( E \right )\leq 1

P(E)=\lim_{n\rightarrow\infty}\left(\frac{r}{n} \right )

 

 

- wherein

Where n repeated experiment and E occurs r times.

 

 

Last place can be  occupied by (0 -9) \:\:\;\;\;\;\;\;10 methods.

  to get '6' at unit place of x^{4} Last digit should be 2, 4 ,6 \:\:or\:\: 8 is  4  ways

\Rightarrow p=\frac{4}{10}=40 \:^{o}/_{o}

Posted by

manish

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE